Graphical Depiction of Reaction (Book Problem 5G.17)






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Rebecca Remple 1C
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Graphical Depiction of Reaction (Book Problem 5G.17)

Postby Rebecca Remple 1C » Sun Mar 01, 2020 2:55 pm

Hi all,

I am currently confused on how to solve 5G.17 in the textbook. I understand that, because the Gibbs Free Energy is positive, I2 increases in pressure while I decreases in pressure. However, I found the answer in the Solutions Manual confusing. Why does the pressure of I2 never equal the pressure of I if they are moving toward equilibrium? Why does the ratio of the reactants to products (1:2) not influence the graph? Also, why is pressure used as the y-axis instead of molar concentration, as was used in Figure 5G.1? If someone could explain these points I would really appreciate it. Thank you!

-Rebecca

Adam Kramer 1A
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Re: Graphical Depiction of Reaction (Book Problem 5G.17)

Postby Adam Kramer 1A » Sun Mar 01, 2020 7:06 pm

The two gases never equal each other because the K constant is greater than one. At the start of the reaction, Q is larger than K, so the backward reaction is favored to go to equilibrium. Because of this, the pressure for 2I is much larger, and it will come down to equilibrium levels as I2 increases to equilibrium levels. Partial Pressure is used as the Y-axis instead of molar concentration because you can use this just as you use concentration to calculate Q and K, so in this case, dealing with gases' partial pressures it makes sense to graph time against partial pressure.

Rebecca Remple 1C
Posts: 137
Joined: Wed Sep 18, 2019 12:16 am
Been upvoted: 1 time

Re: Graphical Depiction of Reaction (Book Problem 5G.17)

Postby Rebecca Remple 1C » Sat Mar 07, 2020 3:45 pm

Adam Kramer 1A wrote:The two gases never equal each other because the K constant is greater than one. At the start of the reaction, Q is larger than K, so the backward reaction is favored to go to equilibrium. Because of this, the pressure for 2I is much larger, and it will come down to equilibrium levels as I2 increases to equilibrium levels. Partial Pressure is used as the Y-axis instead of molar concentration because you can use this just as you use concentration to calculate Q and K, so in this case, dealing with gases' partial pressures it makes sense to graph time against partial pressure.

Hi Adam,

Thank you very much for your explanation! This makes much more sense. However, I was wondering if you could clarify the first part of your explanation. How does K being greater than 1 affect when the reaction reaches equilibrium? I understand how Q being greater than K favors the reverse reaction, but not how K on its own does. I'm not familiar with this concept so I was wondering if you could explain how a large K value affects equilibrium. I would really appreciate it. Thank you again and good luck on the final!

-Rebecca


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